Prove that the equation of the circle ,having double contact with the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$(having eccentricity $e$) at the ends of a latus rectum,is $x^2+y^2-2ae^3x=a^2(1-e^2-e^4)$.
Since the ends of a latus rectum are $(ae,\frac{b^2}{a}),(ae,\frac{-b^2}{a})$.We have to find the equation of circle passing through $(ae,\frac{b^2}{a})$ and ,$(ae,\frac{-b^2}{a})$.I cannot find the equation of circle based on this equation.Any hints/suggestions will help me.Thanks.
HINT... you could find the equation of the normal to the ellipse at $(ae,\frac{b^2}{a})$ and find the point of intersection of this normal and the $x$ axis, which, due to symmetry, would be the centre of the circle